Content-Type: multipart/mixed; boundary="-------------0107020910558" This is a multi-part message in MIME format. ---------------0107020910558 Content-Type: text/plain; name="01-236.keywords" Content-Transfer-Encoding: 7bit Content-Disposition: attachment; filename="01-236.keywords" Absolutely continuous spectrum, Krein systems, Sturm-Liouville operators. ---------------0107020910558 Content-Type: application/x-tex; name="Mnv1.tex" Content-Transfer-Encoding: 7bit Content-Disposition: inline; filename="Mnv1.tex" \documentstyle[12pt]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\textwidth}{7in} \setlength{\textheight}{9in} \setlength{\topmargin}{-.1in} \oddsidemargin=-1cm \language=1 \begin{document} \title{On the existence of the absolutely continuous component for the measure associated with some orthogonal systems. } \date{{\small {\it Moscow State University, Faculty of Computational Mathematics and Cybernetics, Vorob'evy Gory, Moscow, Russia, 119899. e-mail address: saden@cs.msu.su\\}}} \author{Sergey A. Denisov} \maketitle \begin{abstract} In this article, we consider two orthogonal systems: Sturm-Liouville operators and Krein systems. For Krein systems, we study the behavior of generalized polynomials at the infinity for spectral parameters in the upper half-plain. That makes it possible to establish the presence of absolutely continuous component of the associated measure. For Sturm-Liouville operator on the half-line with bounded potential $q$, we prove that essential support of absolutely continuous component of the spectral measure is $[m,\infty)$ if $\limsup_{x\to \infty} q(x)=m$ and $q^\prime\in L^2(R^+)$. That holds for all boundary conditions at zero. This result partially solves one open problem stated recently by S.Molchanov, M.Novitskii, and B.Vainberg. \end{abstract} {\ } \begin{center} {\bf \S 0. Introduction. } \end{center} The contents of the paper is as follows. In the first paragraph, we prove the asymptotics of generalized polynomials for Krein systems with coefficients of a special kind. We also establish the presence of absolutely continuous component of the associated measure. In the second paragraph, results obtained for Krein system are applied to Sturm-Liouville operators. In this introductory paragraph, we will remind some results for the Krein systems. The Krein systems are defined by the equations \begin{equation} \left\{ \begin{array}{cccc} \frac{dP(x,\lambda )}{dx} & = & i\lambda P(x,\lambda )-\overline{A(x)}% P_{*}(x,\lambda ), & \,P(0,\lambda)=1, \\ \frac{dP_{*}(x,\lambda )}{dx} & = & -A(x)P(x,\lambda ), & \, P_*(0,\lambda)=1, \end{array} \right. \label{sys3} \end{equation} where $A(x)$ is locally summable function on $R^+$. In his famous work \cite{Kr}, M.G.Krein showed that $P(x,\lambda)$ have many properties of polynomials orthogonal on the unit circle \footnote{% The spectral theory of Krein systems was developed further in \cite{Akh, Den, Ryb, Sakh1, Sakh2, Sakh3}.}. For example, there exists a non-decreasing function $\sigma(\lambda)$ (spectral measure ) defined on the whole line such that mapping $U_P(f)=\int\limits_0^\infty f(x)P(x,\lambda)dx$ is isometry from $L^2(R^+)$ to $L^2(\sigma,R)$. We will call $P(x,\lambda)$ the generalized polynomials. The following Theorem was stated in \cite{Kr}. It was proved later in \cite{Sakh1}. {\bf Theorem \cite{Kr,Sakh1}.} {\it \ The following statements are equivalent% } {\it (1) The integral $\int\limits_{-\infty }^\infty \frac{\ln \sigma ^{^{\prime }}(\lambda )}{1+\lambda ^2}d\lambda$ is finite.} {\it (2) At least at some }$\lambda $, $\Im \lambda >0$, {\it the integral } \begin{equation} \int\limits_0^\infty \left| P(x,\lambda )\right| ^2dx \label{int3} \end{equation} {\it converges.} {\it (3) At least at some }$\lambda, ${\it \ }$\Im \lambda >0$,{\it \ the function }$P_{*}(x,\lambda )${\it \ is bounded.} {\it (4) On any compact set in the open upper half-plane, integral {\rm (\ref {int3})} converges uniformly. That is equivalent to the existence of uniform limit }$\Pi (\lambda )=\lim_{x\rightarrow \infty }P_{*}(x,\lambda ).${\it \ } Consider some measure $\mu$ on $R$. Let $I$ be the finite union of intervals on $R$. Let us assume that for any measurable set $\Omega\in I$ with positive Lebesgue measure ($|\Omega|>0$ ), we have $\mu (\Omega )>0$. That already means that $\mu$ has nontrivial absolutely continuous component. Though not any measure with nontrivial absolutely continuous component has this property. If this condition holds, we say that the essential support of absolutely continuous component of measure $\mu$ is $I$ ($essupp\{ \mu_{ac} \}=I$). Thus, if one of the conditions {\it (2)-(4)} holds, then the essential support of $\sigma_{ac}(\lambda)$ is $R$. It is easy to show \cite{Kr} that $% A(x)\in L^2(R)$ or $A(x)\in L^1(R)$ yields {\it (3)}. In \cite{Den}, we proved the criterion for {\it (3)} to hold in terms of coefficients $A(x)$ from the so-called Stummel class. In the next paragraph, we will study another class of perturbations that does not shrink the essential support for absolutely continuous component of the measure. The similar problems for Sturm-Liouville operators were studied in numerous publications (see, for example, \cite{MNV} and bibliography there ). But first, let us outline some relations between Krein systems and some other orthogonal systems. Consider the Dirac system \begin{equation} \left\{ \begin{array}{cccc} \phi^\prime & = & -\lambda\psi-a_1\phi+a_2\psi, & \phi(0)=1, \\ \psi^\prime & = & \lambda\phi+a_2\phi+a_1\psi, & \psi(0)=0, \end{array} \right. \label{dir} \end{equation} where $a_1=2\Re A(2x)$, $a_2=2\Im A(2x)$. It turns out that $e^{-i\lambda x}P(2x,\lambda)=\phi(x,\lambda)+i\psi(x,\lambda)$. That allows us to say \cite{Kr} that $\rho_{Dir}(\lambda)=2\sigma(\lambda)$, where $% \rho_{Dir}(\lambda)$-- spectral measure of Dirac systems (\ref{dir}). In case $a_2=0$ and $a_1$-- absolutely continuous, we have \begin{equation} \begin{array}{ccc} \psi^{\prime\prime}-q\psi+\lambda^2\psi=0, & \psi(0)=0, & \psi^\prime(0)=\lambda, \\ \phi^{\prime\prime}-q_1\phi+\lambda^2\phi=0, & \phi(0)=1, & \phi^\prime(0)+a_1(0)\phi(0)=0, \end{array} \end{equation} where $q=a_1^2+a_1^\prime,\ q_1=a_1^2-a_1^\prime$. Therefore, the spectral measure $\rho_d(\lambda)$ of Sturm-Liouville operator $l(u)=-u^{\prime\prime}+qu$ with Dirichlet boundary condition $% u(0)=0$ is related to $\sigma(\lambda)$ by \begin{equation} \rho_d (\lambda)=4\int\limits_0^{\sqrt\lambda} \xi^2d\sigma(\xi),\ \lambda>0. \label{zero} \end{equation} {\ } \begin{center} {\bf \S 1. Krein systems with bounded coefficient which derivative is square summable. } \end{center} In this paragraph, we will prove the following Theorem. {\bf Theorem 1.} {\it If bounded coefficient $A(x)$ of Krein system is real valued,} $02l_2$ is fixed, $k>0 $. Assume for simplicity $A$ is such that $l_1<2A<\tau/2+l_2$ for all $x\in R^+$. We will explain later why this assumption can always be made without loss of generality. Symbol $C$ is reserved for positive constants which value might change from one formula to another. It is easy to verify that $\Re \mu_1>0$ for all $x>0, k>0$. The following Lemma is true. {\bf Lemma.} {\it The asymptotics holds at the infinity } {\it \begin{equation} P_*(x,k)=\exp\left[ \int\limits_0^x \mu_1(s,k) ds \right] O(x,k), \end{equation} } {\it where $|O(x,k)|<\exp(C/k)$ for all $x>0, k>0$}. {\it Proof.} Many estimates in this proof are very crude, but they will be good enough for our purposes. Let $J$ be $2\times2$ matrix that satisfies equation $J^\prime=\aleph J$. We will find $J$ in the form $J=LQ$, where $L=\left[ \begin{array}{cc} -\mu_1/A & -\mu_2/A \\ 1 & 1 \end{array} \right] $ consists of eigenvectors of $\aleph$. We have the following equation for $Q$: $Q^\prime=L^{-1}\aleph LQ-L^{-1}L^\prime Q$. Multiplying matrixes, we have \begin{equation} Q^\prime=\left[ \begin{array}{cc} \mu_1 & 0 \\ 0 & \mu_2 \end{array} \right]Q+VQ, \end{equation} where \begin{equation} V=\left[ \begin{array}{cc} v_{11} & v_{12} \\ v_{21} & v_{22} \end{array} \right] = \frac{1}{\sqrt{4A^2-\lambda^2}} \left[ \begin{array}{cc} -\frac{A^\prime}{A}\mu_1+\mu_1^\prime & -\frac{A^\prime}{A}\mu_2+\mu_2^\prime \\ \frac{A^\prime}{A}\mu_1-\mu_1^\prime & \frac{A^\prime}{A}\mu_2-\mu_2^\prime \end{array} \right]. \end{equation} Let us notice that the following inequality \begin{equation} \displaystyle \int\limits_0^\infty \|V\|^2 dx0$, we have the estimate \begin{equation} \Re \sqrt{4A^2-\lambda^2}<-k. \end{equation} Consequently, Gronwall Lemma, being applied to (\ref{ie}), yields \begin{equation} |s_{11}|\leq \exp\left\{ C \int\limits_0^x |v_{21}(t)| \int\limits_t^\infty |v_{12}(s)| \exp \Bigl( -{k[s-t]} \Bigr) ds dt \right\} \leq \exp(C/k). \end{equation} At the final step, we used (\ref{star}) and the Young inequality for convolutions. Therefore, for $s_{21}$, we have the estimate \[ |s_{21}|\leq \exp(C/k) \int\limits_0^x |v_{21}(s)| |\exp \Bigl[ -\int\limits_0^s \nu(\xi,k) d\xi \Bigr]|ds \] which follows from the second equation of system (\ref{sys1}) and estimate on $s_{11}$. In the same way, the estimates for $s_{12}, s_{22}$ can be obtained. They are as follows \begin{equation} |s_{12}|\leq \frac{C}{\sqrt k} \exp \left( \frac{C}{k} \right), \end{equation} \begin{equation} |s_{22}|\leq \frac{C}{\sqrt k} \exp \left( \frac{C}{k} \right) \int\limits_0^x |v_{21}(s)| |\exp \Bigl[ -\int\limits_0^s \nu(\xi,k) d\xi \Bigr]|ds. \end{equation} If $J$ is such that $J(0)=I$, then $J=LQ_\circ S L^{-1}(0)$. Therefore, for $% P_*$, we will have \begin{equation} P_*=\exp \Bigl[ \int\limits_0^x (\mu_1(t)+v_{11}(t)) dt\Bigr] \Bigl\{ \alpha s_{11}+ \beta s_{12}+\Bigr. \end{equation} \begin{equation} \Bigl. \exp\left( \int\limits_0^x (\mu_2(s)-\mu_1(s)+v_{22}(s)-v_{11}(s))ds \right) (s_{21}\alpha+s_{22}\beta) \Bigr\}. \end{equation} Constants $\alpha$ and $\beta$ are chosen in such a way that the initial condition $P_*(0,k)=1$ is satisfied. We have \begin{equation} \left[ \begin{array}{c} \alpha \\ \beta \end{array} \right] =\frac 1{\sqrt{4A^2(0)-\lambda ^2}}\left( \begin{array}{cc} A(0) & \mu _2(0) \\ -A(0) & -\mu _1(0) \end{array} \right) \times \left[ \begin{array}{c} 1 \\ 1 \end{array} \right] , \end{equation} therefore, $\alpha$ and $\beta$ are bounded uniformly in $k$. Denote by $% O(x,k)$ \begin{equation} O=\exp \Bigl[ \int\limits_0^x v_{11}(t) dt\Bigr] \Bigl\{ \alpha s_{11}+ \beta s_{12}+\Bigr. \label{ooo} \end{equation} \begin{equation} \Bigl. \exp\left( \int\limits_0^x (\mu_2(s)-\mu_1(s)+v_{22}(s)-v_{11}(s))ds \right) (s_{21}\alpha+s_{22}\beta) \Bigr\}. \end{equation} Notice that $\displaystyle \int\limits_0^x v_{11}(t) dt$ and $\displaystyle % \int\limits_0^x v_{22}(t) dt$ are bounded uniformly in $x$ and $k$. Due to the estimates on $s_{21}$ and $s_{22}$, we have \begin{equation} \Bigl| \exp\left( \int\limits_0^x (\mu_2(s)-\mu_1(s)+v_{22}(s)-v_{11}(s))ds \right) (s_{21}\alpha+s_{22}\beta) \Bigr|\leq \label{ff1} \end{equation} \begin{equation} \exp(C/k) \left| \exp \Bigl[ \int\limits_0^x \nu(\eta,k) d\eta \Bigr] \right| \times \end{equation} \begin{equation} \int\limits_0^x |v_{21}(s)| \left| \exp \Bigl[- \int\limits_0^s \nu(\xi,k) d\xi \Bigr] \right| ds \leq \end{equation} \begin{equation} \exp(C/k) \int\limits_0^x |v_{21}(s)| \left| \exp \Bigl[ \int\limits_s^x \nu(\xi,k) d\xi \Bigr] \right| ds \leq \exp(C/k)/{\sqrt k}. \label{ff2} \end{equation} To get the last estimate, we used Cauchy inequality. Finally, bounds on $% s_{11}$, $s_{12}$ lead to $|O(x,k)|<\exp (C/k)$. That finishes the proof of the Lemma. $\Box$ {\it Remark. } More accurate estimates on $\alpha$, $\beta$, $s_{11}$ allow us to write inequality \begin{equation} |O(x,z)-1|<1/2, \label{ret} \end{equation} where $x\in R^+, \tau \leq \Re z \leq \tau +1, \Im z>k_0$, $k_0$-- some positive constant. Indeed, it is easy to verify that $\alpha\to 1$ and $\displaystyle |\beta|<% \frac{C}{\Im z}$ if $\Im z\to +\infty$, $\tau\leq \Re z \leq \tau+1$. From (% \ref{sys1}), by Cauchy inequality we have \begin{equation} |s_{11}-1|\leq \exp(C/k)/k \end{equation} uniformly in $x\in R^+$, $\tau \leq \Re z \leq \tau+1$, where $k=\Im z>k_0$. One can verify that $\displaystyle \int\limits_0^x v_{11}(t) dt\to 0$ uniformly in $x$ if $\Im z\to \infty$. Therefore, from (\ref{ooo}) and (\ref{ff1})-(\ref{ff2}), we infer (\ref{ret}% ). Let us prove Theorem 1 now. {\it Proof of Theorem 1.} Fix any $\tau >2l_2$. We will show that $[\tau ,\infty )\subset essupp\{\sigma _{ac}\}$. Because $\tau $ is chosen arbitrary larger than $2l_2$ and $\sigma $ is odd (since $A$ is real ), this inclusion is sufficient for Theorem 1 to be true. Once $% \tau $ is fixed, we can assume that $l_1<2A<\tau/2 +l_2$ for all $x\in R^{+}$% . Due to the standard trace-class perturbation argument applied to Dirac system (\ref{dir}) \cite{RS3}, we can always make this assumption. Indeed, since $% l_1=\liminf_{x\rightarrow \infty}A\leq \limsup_{x\rightarrow \infty }A=l_2$ as $x\to \infty $, it suffices to multiply $A$ by some smooth function which is equal to $0$ on $[0,M_A]$ and $1$ on $[M_A+1,\infty )$. The absolutely continuous part of $\rho _{Dir}$ will not change because of trace-class argument. On the other hand, if $M_A$ is sufficiently large, we will satisfy the imposed condition. Consider Krein system with coefficients $\displaystyle A_{(n)}(x)=\left\{ \begin{array}{cc} A(x), & x0. \end{array} \right. $ Denote the corresponding measure by $\sigma_n$. We have the formula ((3.14) from [4] ): \begin{equation} \sqrt{2\pi} P_{*}(\infty,z)=\sqrt{2\pi} P_{*}(n,z)= e^{i\alpha_n} \exp \left[ \int\limits_{-\infty}^{\infty} \frac{1}{2\pi i} \frac{(1+tz) \ln \sigma_n^\prime (t)} {(z-t)(1+t^2)}dt \right], \Im z>0, \label{rew} \end{equation} where $\alpha_n$ are some real constants. Because $A(x)$ is real, functions $% \sigma_n$ are odd. Therefore, if we take $z=i$, the left-hand side of (\ref {rew}), together with exponent from the right-hand side, are real valued. Thus, $\alpha_n$ can be chosen equal to zero. The asymptotics of $P_{*}(n,z)$ as $|z|\to \infty$ is $P_{*}(n,\infty)=1$ \cite{Ryb}. Therefore, we can rewrite this formula as follows \begin{equation} -2\pi i \ln P_*(n,z) = \int\limits_{-\infty}^{\infty} \frac{(1+tz) \ln (2\pi \sigma_n^\prime (t))} {(t-z)(1+t^2)}dt \label{ttt} \end{equation} if $\Im z >0$. Here we used the identity \[ \sqrt{2\pi }=\exp \left[ \int\limits_{-\infty}^{\infty} -\frac{1}{2\pi i} \frac{(1+tz) \ln(2\pi) } {(z-t)(1+t^2)}dt \right],\ \Im z>0. \] Also we used the fact that $2\pi \sigma_n^\prime(t)\to 1$ if $|t|\to \infty$ \cite{Ryb}. The right-hand side $RHS(z)$ of (\ref{ttt}) satisfies the condition $% \overline{RHS(z)}=RHS(\bar{z})$. Let us define the left-hand side for $\Im z<0$ according to this rule. So we have some analytic function $LHS(z)$ defined in the region: $\Im z\neq 0$. Recall the asymptotic formula for $% P_*(x,z)$, $(\Im z>0, \Re z=\tau>2l_2 )$: \begin{equation} P_*(x,z)=\exp\left[ \int\limits_0^x \mu_1(s,z) ds \right] O(x,z). \end{equation} Then, \[ LHS(z)=-2\pi i \int\limits_0^n \mu_1(s,z) ds -2\pi i\ln O(n,z) \] for $\Im z>0$. It is easy to see that the continuation according to the chosen rule for function $-2\pi i \int\limits_0^n \mu_1(s,z) ds $ is exactly the Schwarz analytic continuation. It follows from the fact that the value of this function on the half-line $\Im z=0, \Re z\geq \tau$ is real. Consider $z=\tau+ik$. Let $\ln O(n,z)=r_1(n,k)+ir_2(n,k)$. From the definition of $LHS(z)$ in $\Im z<0$, we have $\overline{i(r_1(n,k)+ir_2(n,k)}% =-ir_1(n,k)-r_2(n,k)=ir_1(n,-k)-r_2(n,-k)$. Thus, $r_1(n,k)$ is odd and $% r_2(n,k)$ is even. For fixed $n$, they are well defined functions for all $k \neq 0$ with finite right and left limits at $k=0$. Notice also that for $k>0 $, $r_1(n,k)=\ln |O(n,\tau+ik)|$ and $r_2(n,k)=Arg O(n,\tau+ik)$. For $r_1$, $r_2$,\ we have $r_1(n,\infty)=0$,\ $r_2(n,\infty)=0$ because $% P_*(n,\infty)=1$ and $\mu_1 (s,\infty)=0$. Then, integrate the both sides of (\ref{ttt}), together with some auxiliary function $\displaystyle \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4}$, along the contour $Con$. We choose $Con$ as the contour that consists of the complex numbers $\tau+ik$ ($|k|\leq m-1, |k| \geq m+1$ ) and two right semicircles with radii $1$ centered at $\tau \mp im$. The direction of integration is upward. For the right-hand side, we have \[ \int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \int\limits_{-% \infty}^{\infty} \frac {(1+tz)\ln (2\pi \sigma_n^\prime (t))} {(1+t^2)(t-z)}dt dz= \] \[ \int\limits_{-\infty}^{\infty} \frac{\ln (2\pi \sigma_n^\prime (t))} {1+t^2} \int\limits_{Con} \frac{(1+tz)(z-\tau)^3} {((z-\tau)^2+m^2)^4 (t-z) } dz dt= 2\pi i \int\limits_{\tau}^\infty \frac{(t-\tau)^3} {((t-\tau)^2+m^2)^4} \ln (2\pi \sigma_n^\prime (t))dt. \] Here we changed the order of integration by Fubini Theorem and then used the Cauchy formula. Integrating the $LHS(z)$ with the same function, we have \[ -2\pi i \int\limits_{Con} \int\limits_0^n \mu_1(s,z) ds \frac{(z-\tau)^3}{% ((z-\tau)^2+m^2)^4}dz -2\pi i \int\limits_{Con} \frac{(z-\tau)^3}{% ((z-\tau)^2+m^2)^4} \ln O(n,z)dz= \] \[ 0-2\pi i \int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \ln O(n,z)dz \] because $\int\limits_0^n \mu_1(s,z) ds$ is analytic and bounded in $\Re z \geq \tau$. Thus, we have the equality \begin{equation} - \int\limits_{Con} \frac{(z-\tau)^3}{((z-\tau)^2+m^2)^4} \ln O(n,z) dz= \int\limits_{\tau}^\infty \frac{(t-\tau)^3}{((t-\tau)^2+m^2)^4} \ln (2\pi \sigma_n^\prime (t))dt. \label{ger} \end{equation} We have the following inequality \cite{Sakh1} \begin{equation} \int\limits_{-\infty}^\infty \frac{d\sigma_n(t)}{1+t^2}1$ and zero otherwise, $% \ln^- h= -\ln h$ if $0m+1} i\frac {(ik)^3}{(m^2-k^2)^4} (r_1(n,k)+ir_2(n,k))dk= \] \[ 2\int\limits_{0m+1} \frac{k^3}{(m^2-k^2)^4} r_1(n,k)dk \leq C \] uniformly in $n$ due to the estimates on $\ln |O(n,k)|$. We also used the fact that $r_2(n,k)$ is even. It is crucial since we do not have control over the argument of $O(n,k)$. We only know the upper bound on $\ln |O(n,k)|$% . Term $I_2$ corresponds to the integration along the semicircles. Choose $m$ sufficiently large. Then, for $I_2$, we have the estimate \[ |I_2|\leq C \] uniformly in $n$ because of the estimate (\ref{ret}). To finish the proof, we will use one argument from \cite{DK}. Consider any compact set $Co\in (\tau, \infty)$ of positive Lebesgue measure. Then, as it follows from (\ref{pp}) and boundedness of right-hand side in (\ref{ger}), $% \int\limits_{Co} \ln^- \sigma_n^\prime (t) dt$ is bounded uniformly in $n$. Jensen's inequality yields \[ \ln^- \left\{ \frac{1}{|Co|}\int\limits_{Co} \sigma_n^\prime(t)dt \right\} \leq \frac{1}{|Co|} \int\limits_{Co} \ln^- \sigma_n^\prime (t)dt. \] Therefore, $\sigma_n(Co)$ is greater than some positive constant $d(Co)$ for all $n$. The Weyl-Titchmarsh function of system with coefficients $A_{(n)}$ converges to Weyl-Titchmarsh function of system with coefficient $A$ \cite {Sakh1}. This convergence is uniform on any compact set of upper half-plain. By Stone-Weierstrass Theorem, we have weak convergence of $\sigma_n$ to $% \sigma$. Consequently, $\sigma(Co)\geq \limsup _{n\to \infty} \sigma_n(Co) >0 $ for each compact $Co$ with $|Co|>0$. $\Box$ {\ } \begin{center} {\bf \S 2. Sturm-Liouville operators. } \end{center} The main goal of this paragraph is to prove the following Theorem. The idea of the proof will follow \cite{Den1} more or less. But it will require more technical details. {\bf Theorem 2. } {\it Consider Sturm-Liouville operator on the half-line given by differential expression } {\it \begin{equation} l(u)=-u^{\prime\prime}+qu, \label{sl} \end{equation} } {\it where $q$ is bounded, }$\limsup_{x\rightarrow \infty }q=m,${\it \ $q(x)$ is absolutely continuous, and $q^{\prime }\in L^2(R^{+})$. Then, for any boundary condition at zero, the essential support of absolutely continuous component of the corresponding spectral measure is }$[m,\infty )${\it .} The problems of this kind for Sturm-Liouville operators were treated in many publications. We mention some of the results obtained. P.Deift and R.Killip proved {\bf Theorem \cite{DK}.} {\it If in (\ref{sl}), $q(x)\in L^2(R^+)$, then the essential support of absolutely continuous component of spectral measure is $% R^+$ for all boundary conditions.} In the paper of S.Molchanov, M.Novitskii, and B.Vainberg \cite{MNV}, some other results were obtained. For example, it was proved that $q\in L^3(R^{+}) $ and $q^{\prime }\in L^2(R^{+})$ lead to the same property of the spectral measure. Consider the particular case of Theorem 2, where $q\to 0$ as $x\to\infty$, $q^{\prime}\in L^2(R^+)$. We see that condition $q\in L^3(R^+)$ used in \cite{MNV} can be relaxed to $q\to 0$ as $x\to \infty$. In \cite{MNV}, authors also pose the following open problem. Is it true that for all boundary conditions at zero, $essupp \{ \rho_{ac} \} =[m,\infty )\ $ provided that $q$ is bounded, $\ \limsup_{x\rightarrow \infty }q=m$, and for some integer $p$, $q^{(p)}\in L^2(R^{+})$. Here $q^{(p)}$ means the derivative of order $p$. Theorem 2 solves this problem for $p=1$. Because any bounded function admits supremum, we actually characterize the absolutely continuous component of the spectrum for bounded potentials with square summable first derivative. We think that method developed here can be used to deal with any $p$. No doubts, it will require more calculations to establish the asymptotics. Let us prove the following Lemma first. {\bf Lemma.} {\it If bounded }$q${\it \ is such that }$\limsup_{x\rightarrow \infty }q=m${\it \ and $q^{\prime }\in L^2(R^{+})$, then for some large $% \gamma >0$, there exists bounded $v(x)$ so that }$\limsup_{x\rightarrow \infty }v=\sqrt{\gamma ^2+m}-\gamma $, {\it \ $v^{\prime }\in L^2(R^{+})$, and $q=v^2+2\gamma v+v^{\prime }$.} {\it Proof }. Consider the corresponding integral equation \begin{equation} v(x)=e^{-2\gamma x} \int\limits_0^x e^{2\gamma s} (q(s)-v^2(s))ds. \label{m1} \end{equation} If $v(x)$ is solution of this integral equation, then it satisfies differential equation as well. Write (\ref{m1}) as follows $v=OP_\gamma v$, where $OP_\gamma $ is corresponding formal nonlinear operator. Consider the complete metric space $M=\{f\in C(R^{+}),\Vert f\Vert _\infty \leq 1\}$ with $\Vert .\Vert _\infty $ metric. By $C(R^{+})$ we denote continuous functions on $R^{+}$. If $\gamma $ is large enough, operator $OP_\gamma $ acts from $M$ to $M$. Naturally, the choice of $\gamma$ depends on $\|q\|_\infty$. For large $\gamma $, $OP_\gamma $ has contracting property. Indeed, \[ |OP_\gamma g_1 -OP_\gamma g_2 | \leq e^{-2\gamma x} \int\limits_0^x e^{2\gamma s} |g_1-g_2| |g_1+g_2|ds\leq \frac 12 \|g_1-g_2\|_\infty, \] for $\gamma$ large enough. Therefore, there is the unique fixed point from $% M $. Let us call this function $v$. Differentiate ($\ref{m1}$). After integration by parts, we will have \[ v^\prime =e^{-2\gamma x}(q(0)-v^2(0))+e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} q^\prime (s)ds-2e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} v(s)v^\prime (s)ds. \] Consider integral equation \begin{equation} b =e^{-2\gamma x}(q(0)-v^2(0))+e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} q^\prime (s)ds-2e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} v(s)b(s)ds. \label{m2} \end{equation} It has the unique solution from $C(R^+)$. Therefore, its solution is $% v^\prime$. The uniqueness follows from the convergence of the corresponding iterated series. Write (\ref{m2}) as $b=OP1_\gamma b$. \[ e^{-2\gamma x}\int\limits_0^x e^{2\gamma s} q^\prime (s)ds=\int_0^\infty r_\gamma (x-s) q^\prime(s)ds, \] where $r_\gamma(s)=e^{-2\gamma s}$ for positive $s$, and zero otherwise. Because $q^\prime\in L^2(R^+)$, the Young inequality for convolutions yields that $OP1_\gamma$ acts from ball $\|.\|_2\leq 1$ into itself, provided that $% \gamma$ is large enough. For large $\gamma$, it has contracting property. Therefore, there is the unique fixed point $b$ in ball $\|.\|_2\leq 1$ which is equal to $v^\prime$. We also used the fact that $\|v\|_\infty \leq 1 $. It is clear that $v^\prime (x)\to 0$ as $x \to \infty$. Therefore, solving equation $v^2+2\gamma v+v^\prime-q=0$ gives formula \begin{equation} v=-\gamma \pm \sqrt{\gamma^2 -v^\prime +q}. \label{ame} \end{equation} We know that $q$ is bounded and $\|v\|_\infty\leq 1$ for large $\gamma>0$. Consequently, to obtain asymptotics of $v$ at the infinity, one should take sign $+$ in (\ref{ame}). Therefore, we have $\limsup_{x\to \infty} v(x) =-\gamma+\sqrt{\gamma^2+m}$. $\Box$ Now the proof of Theorem 2 is straightforward. {\it Proof of the Theorem 2.} Notice that the essential support of the absolutely continuous component does not depend on the boundary condition at zero. It follows, for example, from the subordinacy theory \cite{GP,Sim}. Consider (\ref{sl}) with Dirichlet boundary condition at zero. Add to the potential $q$ some constant $\gamma ^2$. Denote the corresponding self-adjoint operator by $H_\gamma $. Obviously, the spectral measure of $H_\gamma $ is the shift of the spectral measure of the initial operator $H_0$. On the other hand, for large $\gamma $, we can solve equation $q+\gamma ^2=(v+\gamma )^2+(v+\gamma )^{\prime }$. That is due to Lemma of this paragraph. Now we can apply results of the first paragraph for Krein systems with coefficient $% A(x)=\gamma /2+v(x/2)/2$. We can use Theorem 1 because for large $\gamma$, $% \displaystyle 0<\liminf_{x\to \infty} A(x) \leq \limsup_{x\to \infty} A(x) = \frac{\sqrt{\gamma^2+m}}{2}$. Let us use formula (\ref{zero}) from the Introduction. Thus, we see that $[\gamma ^2+m,\infty )\subset essupp \{\rho _{ac}(H_\gamma )\}$ which is equivalent to $[m,\infty )\subset essupp \{\rho _{ac}(H_0)\}$. To prove that it is actually an equality, one should use some result from \cite{St} which goes back to \cite{SS}. Before formulating this result, let us introduce some notations used in \cite{St}. Consider $V$ -- real valued and locally integrable on $(0,\infty ).$ Assume that $-\frac{d^2}{dx^2}+V$ is limit point at $+\infty $ and $\int\limits_0^\varepsilon |V(s)|ds<\infty . $ Let $T=-\frac{d^2}{dx^2}+V$ in $L^2(R^{+})$ with any fixed boundary condition at zero. Consider also $V_{\circ }$ -- integrable and bounded from below function. Denote by $T_{\circ }$ operator generated by $-\frac{d^2}{% dx^2}+V_{\circ }$ in $L^2(R).$ Then, the following theorem holds. {\bf Theorem (\cite{St}, Theorem 2.1 ).} {\it Assume that }$-\infty \leq \alpha <\beta \leq +\infty \ ${\it and }$(\alpha ,\beta )\cap Spectrum (T_{\circ })=\emptyset \ ${\it and that intervals }$I_n\subset [0,\infty )$% {\it \ exist such that } \[ |I_n|\rightarrow \infty \ as\ n\rightarrow \infty \ and\ \sup_{x\in \cup _nI_n}|V(x)-V_{\circ }(x)|\leq \delta . \] {\it Then, } $essupp \{ \rho _{ac}(T) \} \cap (\alpha +\delta ,\beta -\delta )=\emptyset .$ Now, consider $V_{\circ }=m$, $V=q$. Because $\limsup_{x\to \infty }q=m$ and $q^{\prime }\in L^2(R^{+})$, one can easily show that for any $\delta >0$, $% essupp \{ \rho _{ac}(H_0) \} \cap (-\infty ,m-\delta )=\emptyset $. Indeed, it suffices to choose $x_n\to \infty $ such that $|q(x_n)-m|<\delta/2 $ and $% I_n$ as some neighboring intervals. We specify them as follows. Notice that $% q(x)-q(x_n)=\int\limits_{x_n}^x q^{\prime }(s)ds$. Therefore, $|q(x)-q(x_n)|\leq \left( \int\limits_{x_n}^\infty \left[ q^{\prime }(s)\right] ^2ds\right) ^{1/2}\cdot \sqrt{x-x_n}=\varepsilon _n\sqrt{x-x_n}, $ where $\varepsilon _n\rightarrow 0$ as $n\rightarrow \infty $. Choose as $% I_n $ intervals $[x_n,x_n+\varepsilon _n^{-1}]$ for $n$ so large that $% \varepsilon_n<\frac{\delta^2}{4}$. Evidently, $|I_n|\rightarrow \infty $ and for each $x\in I_n$, we have $|q(x_n)-m|<\delta $. Therefore, Stolz Theorem yields $essupp \{\rho _{ac}(H_0)\}=[m,\infty ).\Box $ {\it Remark.} There are many functions that satisfy conditions of Theorem 2. In particular, these are some slowly oscillating functions. One can think about $\cos (x^\mu )$ for $0<\mu <1/2$. In that case $\limsup_{x\to \infty }q(x)=1$, $q^{\prime }\in L^2(R^{+})$. Consequently, $essupp \{ \rho _{ac} \} =[1,\infty )$ for all boundary conditions. In paper \cite{St}, author used the theory of subordinate solutions to show that for any $0<\mu <1$, the spectrum is actually purely absolutely continuous on $[1,\infty )$. The interval $[-1,1]$ is filled by singular spectrum. {\it Remark.} The condition $q^{\prime }\in L^2(R^{+})$ from Theorem 2 is optimal \cite{MNV}. That means that the statement can be false if $% q^{\prime}\in L^p(R^{+}),\ p>2$. Famous von Neumann-Wigner potential \cite {vNW,RS4} satisfies the conditions of the Theorem 2. That means that under the conditions of Theorem 2, the singular component of spectrum can appear on the interval $[m,\infty )$ which supports the absolutely continuous part. {\it Remark.} It should be noted that class of Sturm-Liouville operators with decaying potentials match very well the class of Krein systems with coefficients that tend to nonzero constant. For example, Sturm-Liouville operators of this kind admit negative eigenvalues with zero as the only possible point of accumulation. For the Krein systems, these eigenvalues of the discrete spectrum might accumulate only near the edges of the corresponding symmetric interval centered at zero. This relation can be explained as follows. Consider potential from $L^p(R^+), (1\leq p<\infty )$ space and Dirichlet boundary condition for example. Then, for $\gamma$ large enough, equation (\ref{m1}) can be solved so that $v\in L^\infty (R^+)\cap L^p(R^+)$. Consequently, formula (\ref{zero}) from the Introduction allows us to study Krein system with coefficient $A=\gamma/2+v(x/2)/2$ instead of initial Sturm-Liouville operator. Methods, developed in this paper, can be applied to some other classes of Sturm-Liouville operators and Krein systems. For example, using the same arguments, one can prove that the result of P.Deift and R.Killip holds for more general class of potentials. Namely, for uniformly square summable functions from $H^{-1}(R^+)$ space, i.e. \[ \int\limits_x^{x+1} q^2(s)ds < C \] uniformly in $x\in R^+$, and \[ e^{-x} \int\limits_0^x e^s q(s)ds \in L^2(R^+). \] That allows to treat some oscillations at the infinity. This result solves one open problem stated in \cite{Den1}. The proof is essentially the same as the proof of Theorem 2. Instead of studying the Sturm-Liouville operator, we consider the corresponding Krein system with coefficient $A(x)=\gamma/2+v(x/2)/2$, where bounded $v(x)$ is from $L^2(R^+)$. The methods from Theorem 1 are applied then. {\it Acknowledgment.} Author thanks professor J.Goodman for useful discussion. Most of this work was done during the stay at Courant Institute of Mathematical Sciences. 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